Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 1498 by Rasheed Soomro last updated on 14/Aug/15

Find complex numbers α and β such that α^( m) =β^( n) and β^( m) =α^n , m,n ∈ Z Determine formula for the number of pairs (α,β) fulfilling the above conditions. ( You may ignore this part in your answer)

$$\mathrm{Find}\:\mathrm{complex}\:\mathrm{numbers}\:\alpha\:{and}\:\:\beta\:\:\mathrm{such}\:\mathrm{that} \\ $$$$\alpha^{\:{m}} =\beta^{\:\:{n}} \:\:\:\:{and}\:\:\:\beta^{\:\:{m}} =\alpha^{\mathrm{n}} \:\:,\:{m},{n}\:\in\:\mathbb{Z} \\ $$$$\boldsymbol{\mathrm{D}}\mathrm{etermine}\:\mathrm{formula}\:\mathrm{for}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{pairs}\:\left(\alpha,\beta\right)\:\mathrm{fulfilling}\:\mathrm{the} \\ $$$$\mathrm{above}\:\mathrm{conditions}.\:\:\left(\:\mathrm{You}\:\mathrm{may}\:\mathrm{ignore}\:\mathrm{this}\:\mathrm{part}\:\mathrm{in}\:\mathrm{your}\:\mathrm{answer}\right) \\ $$

Commented by 123456 last updated on 14/Aug/15

(n,m)≠(0,0)⇒(α,β)=(0,0) is one of the solutions

$$\left({n},{m}\right)\neq\left(\mathrm{0},\mathrm{0}\right)\Rightarrow\left(\alpha,\beta\right)=\left(\mathrm{0},\mathrm{0}\right)\:\mathrm{is}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solutions} \\ $$

Commented by Rasheed Ahmad last updated on 14/Aug/15

An other pair is (1,1) but these are real solutions only.

$$\mathrm{An}\:\mathrm{other}\:\mathrm{pair}\:\mathrm{is}\:\left(\mathrm{1},\mathrm{1}\right)\:{but}\:{these}\:{are} \\ $$$${real}\:{solutions}\:{only}. \\ $$

Commented by 123456 last updated on 15/Aug/15

{ ((α^m =β^n )),((β^m =α^n )) :}⇒_(β→α) ^(α→β) { ((β^m =α^n )),((α^m =β^n )) :} (α,β) is a solution, then (β,α) is also a solution

$$\begin{cases}{\alpha^{{m}} =\beta^{{n}} }\\{\beta^{{m}} =\alpha^{{n}} }\end{cases}\underset{\beta\rightarrow\alpha} {\overset{\alpha\rightarrow\beta} {\Rightarrow}}\begin{cases}{\beta^{{m}} =\alpha^{{n}} }\\{\alpha^{{m}} =\beta^{{n}} }\end{cases} \\ $$$$\left(\alpha,\beta\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution},\:\mathrm{then}\:\left(\beta,\alpha\right)\:\mathrm{is}\:\mathrm{also}\:\mathrm{a}\:\mathrm{solution} \\ $$

Commented by Rasheed Ahmad last updated on 15/Aug/15

Good deduction!

$$\mathrm{Good}\:\mathrm{deduction}! \\ $$

Commented by 123456 last updated on 15/Aug/15

(n,m)∈(N^∗ )^2 ,p≤max(rn,rm)+1 k=lcm(n,m) { ((α^n =β^m )),((β^n =α^m )) :}⇒ { ((α^(nk_1 ) =β^(mk_1 ) )),((β^(nk_2 ) =α^(mk_2 ) )) :} mk_1 =nk_2 =k α^(nk_1 ) −α^(mk_2 ) =0⇒α^r_1 (α^(r_2 −r_1 ) −1)=0 r_1 =min(nk_1 ,mk_2 ),r_2 =max(nk_1 ,mk_2 ) α^r_1 =0∨α^r =1 r=r_2 −r_1

$$\left({n},{m}\right)\in\left(\mathbb{N}^{\ast} \right)^{\mathrm{2}} ,{p}\leqslant\mathrm{max}\left({rn},{rm}\right)+\mathrm{1} \\ $$$${k}=\mathrm{lcm}\left({n},{m}\right) \\ $$$$\begin{cases}{\alpha^{{n}} =\beta^{{m}} }\\{\beta^{{n}} =\alpha^{{m}} }\end{cases}\Rightarrow\begin{cases}{\alpha^{{nk}_{\mathrm{1}} } =\beta^{{mk}_{\mathrm{1}} } }\\{\beta^{{nk}_{\mathrm{2}} } =\alpha^{{mk}_{\mathrm{2}} } }\end{cases} \\ $$$${mk}_{\mathrm{1}} ={nk}_{\mathrm{2}} ={k} \\ $$$$\alpha^{{nk}_{\mathrm{1}} } −\alpha^{{mk}_{\mathrm{2}} } =\mathrm{0}\Rightarrow\alpha^{{r}_{\mathrm{1}} } \left(\alpha^{{r}_{\mathrm{2}} −{r}_{\mathrm{1}} } −\mathrm{1}\right)=\mathrm{0} \\ $$$${r}_{\mathrm{1}} =\mathrm{min}\left({nk}_{\mathrm{1}} ,{mk}_{\mathrm{2}} \right),{r}_{\mathrm{2}} =\mathrm{max}\left({nk}_{\mathrm{1}} ,{mk}_{\mathrm{2}} \right) \\ $$$$\alpha^{{r}_{\mathrm{1}} } =\mathrm{0}\vee\alpha^{{r}} =\mathrm{1} \\ $$$${r}={r}_{\mathrm{2}} −{r}_{\mathrm{1}} \\ $$

Commented by Rasheed Soomro last updated on 15/Aug/15

I have used your this result in my answer to Q 1497

$${I}\:{have}\:{used}\:{your}\:{this}\:{result}\:{in}\:{my}\:{answer}\:{to}\:{Q}\:\mathrm{1497} \\ $$

Answered by Rasheed Soomro last updated on 17/Aug/15

Let gcd(m,n)=k ⇒ lcm(m,n)=((mn)/k) α^m =β^( n) ⇒( α^m )^(m/k) =(β^( n) )^(m/k) ⇒ α^(m^2 /k) =β^((mn)/k) (I) β^( m) =α^( n) ⇒ (β^( m) )^(n/k) = (α^n )^(n/k) ⇒ β^((mn)/k) =α^(n^2 /k) (II) Our goal is to achieve β ′s common exponent ( of course least common exponent (((mn)/k) ) is suitable for simplified solution) in order to eliminate β in above steps. From (I) and (II) α^(m^2 /k) = α^(n^2 /k) As m and n are exchangeable here we may assume m≥n (If n>m then it will be merely matter of exchanging symbols) α^(m^2 /k) − α^(n^2 /k) =0 ⇒ α^(n^2 /k) ( α^((m^2 /k) − (n^2 /k)) − 1)=0 ⇒ α^(n^2 /k) = 0 ∨ α^((m^2 − n^2 )/k) =1 ⇒ α=0 ∨ α=^((m^2 − n^2 )/k) (√( 1)) (α is (((m^2 − n^2 )/k))th root of unity) α=0 ⇒β=0 [ No need to show process ] ∴ α is (((m^2 − n^2 )/k))th root of unity. Similarly it can be shown that, β is also (((m^2 − n^2 )/k))th root of unity. nth roots of unity are given by: ( cos ((2π)/n) +ı sin ((2π)/n) )^j j=0,1,2,....,n−1 Hence, (((m^2 − n^2 )/k)) th roots of unity will be: ( cos ((2πk)/(m^2 −n^2 )) +ı sin ((2πk)/(m^2 −n^2 )) )^j j=0,1,2,...(.((m^2 − n^2 )/k) −1) Let cos ((2πk)/(m^2 −n^2 )) +ı sin ((2πk)/(m^2 −n^2 )) =ω (((m^2 − n^2 )/k)) th roots of unity=1,ω,ω^2 ,......,ω^((m^2 − n^2 −k)/k) [ Let ((m^2 −n^2 )/k) = N ] mth power of ∽ =1, ω^( m(mod N)) ,ω^( 2m(mod N)) ,...ω^( m(N−1)(mod N)) nth power of ∽ =1, ω^( n(mod N)) ,ω^( 2n(mod N)) ,...ω^( n(N−1)(mod N)) Let j_1 , j_2 =0,1,2,...(.((m^2 − n^2 )/k) −1) Let α = ω^( j_1 ) and β = ω^( j_2 ) are two Nth roots such that α^m =β^( n) ⇒ ω^( mj_1 (mod N)) =ω^( nj_2 (mod N)) ⇒mj_1 (mod N)=nj_2 (mod N) α^n =β^( m) ⇒ ω^( nj_1 (mod N)) =ω^( mj_2 (mod N)) ⇒nj_1 (mod N)=mj_2 (mod N) Therefore −−−−−−−−−−−−−−−−−−−−−−−−− For mj_1 (mod ((m^2 −n^2 )/k) )=nj_2 (mod ((m^2 −n^2 )/k) ) and nj_1 (mod ((m^2 −n^2 )/k) )=mj_2 (mod ((m^2 −n^2 )/k) ) ( α, β)=(ω^j_1 , ω^j_2 )=(ω^j_2 , ω^j_1 ) where ω is (((m^2 − n^2 )/k)) th root of unity( with m≥n).

$$\:\:\:\:\:\mathrm{Let}\:\mathrm{gcd}\left(\mathrm{m},{n}\right)=\mathrm{k}\:\Rightarrow\:\:\mathrm{lcm}\left(\mathrm{m},{n}\right)=\frac{\mathrm{mn}}{\mathrm{k}} \\ $$$$\alpha^{{m}} =\beta^{\:{n}} \:\Rightarrow\left(\:\alpha^{{m}} \right)^{\frac{\mathrm{m}}{\mathrm{k}}} =\left(\beta^{\:\mathrm{n}} \right)^{\frac{\mathrm{m}}{\mathrm{k}}} \Rightarrow\:\alpha^{\frac{\mathrm{m}^{\mathrm{2}} }{\mathrm{k}}} =\beta^{\frac{\mathrm{mn}}{\mathrm{k}}} \:\:\:\:\:\:\:\:\:\left(\mathrm{I}\right) \\ $$$$\beta^{\:\mathrm{m}} =\alpha^{\:\mathrm{n}} \:\Rightarrow\:\left(\beta^{\:\mathrm{m}} \right)^{\frac{\mathrm{n}}{\mathrm{k}}} =\:\left(\alpha^{\mathrm{n}} \right)^{\frac{\mathrm{n}}{\mathrm{k}}} \Rightarrow\:\beta^{\frac{\mathrm{mn}}{\mathrm{k}}} =\alpha^{\frac{\mathrm{n}^{\mathrm{2}} }{\mathrm{k}}} \:\:\:\:\:\:\:\:\:\left(\mathrm{II}\right) \\ $$$$\mathrm{Our}\:\mathrm{goal}\:\mathrm{is}\:\mathrm{to}\:\mathrm{achieve}\:\beta\:'\mathrm{s}\:\:\mathrm{common}\:\mathrm{exponent}\: \\ $$$$\:\left(\:\mathrm{of}\:\mathrm{course}\:\:\mathrm{least}\:\mathrm{common}\:\mathrm{exponent}\:\left(\frac{\mathrm{mn}}{\mathrm{k}}\:\right)\:\mathrm{is}\:\:\mathrm{suitable}\:\mathrm{for}\:\mathrm{simplified}\:\mathrm{solution}\right)\: \\ $$$$\:\:\:\mathrm{in}\:\mathrm{order}\:\mathrm{to}\:\mathrm{eliminate}\:\beta\:\:\:\mathrm{in}\:\mathrm{above}\:\mathrm{steps}. \\ $$$$ \\ $$$$\mathrm{From}\:\left(\mathrm{I}\right)\:\mathrm{and}\:\left(\mathrm{II}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\alpha^{\frac{\mathrm{m}^{\mathrm{2}} }{\mathrm{k}}} =\:\alpha^{\frac{\mathrm{n}^{\mathrm{2}} }{\mathrm{k}}} \: \\ $$$$\mathrm{As}\:\mathrm{m}\:\mathrm{and}\:\mathrm{n}\:\mathrm{are}\:\mathrm{exchangeable}\:\mathrm{here}\:\mathrm{we}\:\mathrm{may}\:\mathrm{assume}\:\mathrm{m}\geqslant{n}\: \\ $$$$\left(\mathrm{If}\:\mathrm{n}>\mathrm{m}\:\mathrm{then}\:\mathrm{it}\:\mathrm{will}\:\mathrm{be}\:\:\mathrm{merely}\:\mathrm{matter}\:\mathrm{of}\:\mathrm{exchanging}\:\mathrm{symbols}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\alpha^{\frac{\mathrm{m}^{\mathrm{2}} }{\mathrm{k}}} −\:\alpha^{\frac{\mathrm{n}^{\mathrm{2}} }{\mathrm{k}}} \:=\mathrm{0}\:\Rightarrow\:\alpha^{\frac{\mathrm{n}^{\mathrm{2}} }{\mathrm{k}}} \left(\:\alpha^{\frac{\mathrm{m}^{\mathrm{2}} }{\mathrm{k}}\:−\:\frac{\mathrm{n}^{\mathrm{2}} }{\mathrm{k}}} −\:\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\Rightarrow\:\alpha^{\frac{\mathrm{n}^{\mathrm{2}} }{\mathrm{k}}} =\:\mathrm{0}\:\vee\:\alpha^{\frac{\mathrm{m}^{\mathrm{2}} −\:\mathrm{n}^{\mathrm{2}} }{\mathrm{k}}} =\mathrm{1} \\ $$$$\:\:\:\:\:\Rightarrow\:\alpha=\mathrm{0}\:\:\vee\:\alpha=^{\frac{\mathrm{m}^{\mathrm{2}} \:−\:\mathrm{n}^{\mathrm{2}} }{\mathrm{k}}} \sqrt{\:\mathrm{1}}\:\:\:\:\left(\alpha\:\mathrm{is}\:\left(\frac{\mathrm{m}^{\mathrm{2}} \:−\:\mathrm{n}^{\mathrm{2}} }{\mathrm{k}}\right)\mathrm{th}\:\mathrm{root}\:\mathrm{of}\:\mathrm{unity}\right) \\ $$$$\alpha=\mathrm{0}\:\Rightarrow\beta=\mathrm{0}\:\:\:\:\left[\:\mathrm{No}\:\mathrm{need}\:\mathrm{to}\:\mathrm{show}\:\mathrm{process}\:\right] \\ $$$$\:\:\:\:\therefore\:\:\:\:\:\:\alpha\:\mathrm{is}\:\left(\frac{\mathrm{m}^{\mathrm{2}} \:−\:\mathrm{n}^{\mathrm{2}} }{\mathrm{k}}\right)\mathrm{th}\:\mathrm{root}\:\mathrm{of}\:\mathrm{unity}. \\ $$$$\mathrm{Similarly}\:\mathrm{it}\:\mathrm{can}\:\mathrm{be}\:\mathrm{shown}\:\mathrm{that}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\beta\:\:\mathrm{is}\:\mathrm{also}\:\:\left(\frac{\mathrm{m}^{\mathrm{2}} \:−\:\mathrm{n}^{\mathrm{2}} }{\mathrm{k}}\right)\mathrm{th}\:\mathrm{root}\:\mathrm{of}\:\mathrm{unity}. \\ $$$$\:\:\:\mathrm{nth}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{unity}\:\mathrm{are}\:\mathrm{given}\:\mathrm{by}:\: \\ $$$$\:\:\:\:\:\:\:\:\left(\:\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{n}}\:+\imath\:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{n}}\:\:\right)^{\mathrm{j}} \:\:\:\:\:\mathrm{j}=\mathrm{0},\mathrm{1},\mathrm{2},....,\mathrm{n}−\mathrm{1} \\ $$$$\mathrm{Hence}, \\ $$$$\:\:\:\:\:\:\:\left(\frac{\mathrm{m}^{\mathrm{2}} \:−\:\mathrm{n}^{\mathrm{2}} }{\mathrm{k}}\right)\:{th}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{unity}\:\mathrm{will}\:\mathrm{be}: \\ $$$$\:\:\:\:\:\left(\:\mathrm{cos}\:\frac{\mathrm{2}\pi\mathrm{k}}{\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} }\:+\imath\:\mathrm{sin}\:\frac{\mathrm{2}\pi\mathrm{k}}{\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} }\:\right)^{\mathrm{j}} \:\:\:\:\:\:\:\mathrm{j}=\mathrm{0},\mathrm{1},\mathrm{2},...\left(.\frac{\mathrm{m}^{\mathrm{2}} \:−\:\mathrm{n}^{\mathrm{2}} }{\mathrm{k}}\:−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\mathrm{Let}\:\:\mathrm{cos}\:\frac{\mathrm{2}\pi\mathrm{k}}{\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} }\:+\imath\:\mathrm{sin}\:\frac{\mathrm{2}\pi\mathrm{k}}{\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} }\:\:\:\:=\omega \\ $$$$\left(\frac{\mathrm{m}^{\mathrm{2}} \:−\:\mathrm{n}^{\mathrm{2}} }{\mathrm{k}}\right)\:{th}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{unity}=\mathrm{1},\omega,\omega^{\mathrm{2}} ,......,\omega^{\frac{{m}^{\mathrm{2}} −\:\mathrm{n}^{\mathrm{2}} −\mathrm{k}}{\mathrm{k}}} \\ $$$$\left[\:\mathrm{Let}\:\frac{\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} }{\mathrm{k}}\:=\:\mathrm{N}\:\right] \\ $$$$\mathrm{m}{th}\:\mathrm{power}\:\mathrm{of}\:\:\:\:\backsim\:\:\:=\mathrm{1},\:\omega^{\:\mathrm{m}\left(\mathrm{mod}\:\mathrm{N}\right)} ,\omega^{\:\mathrm{2m}\left(\mathrm{mod}\:\mathrm{N}\right)} ,...\omega^{\:\mathrm{m}\left(\mathrm{N}−\mathrm{1}\right)\left(\mathrm{mod}\:\mathrm{N}\right)} \\ $$$$\mathrm{n}{th}\:\mathrm{power}\:\mathrm{of}\:\:\:\:\:\backsim\:\:\:\:=\mathrm{1},\:\omega^{\:\mathrm{n}\left(\mathrm{mod}\:\mathrm{N}\right)} ,\omega^{\:\mathrm{2n}\left(\mathrm{mod}\:\mathrm{N}\right)} ,...\omega^{\:\mathrm{n}\left(\mathrm{N}−\mathrm{1}\right)\left(\mathrm{mod}\:\mathrm{N}\right)} \\ $$$$\mathrm{Let}\:\:\mathrm{j}_{\mathrm{1}} \:,\:\mathrm{j}_{\mathrm{2}} =\mathrm{0},\mathrm{1},\mathrm{2},...\left(.\frac{\mathrm{m}^{\mathrm{2}} \:−\:\mathrm{n}^{\mathrm{2}} }{\mathrm{k}}\:−\mathrm{1}\right) \\ $$$$\mathrm{Let}\:\alpha\:=\:\omega^{\:{j}_{\mathrm{1}} } \:\:\mathrm{and}\:\beta\:=\:\:\omega^{\:\mathrm{j}_{\mathrm{2}} } \:\:\mathrm{are}\:\mathrm{two}\:\mathrm{N}{th}\:\mathrm{roots}\:\mathrm{such}\:\mathrm{that} \\ $$$$\:\:\alpha^{\mathrm{m}} =\beta^{\:\mathrm{n}} \Rightarrow\:\:\:\omega^{\:\mathrm{mj}_{\mathrm{1}} \left(\mathrm{mod}\:\mathrm{N}\right)} =\omega^{\:\mathrm{nj}_{\mathrm{2}} \left(\mathrm{mod}\:\mathrm{N}\right)} \Rightarrow\mathrm{mj}_{\mathrm{1}} \left(\mathrm{mod}\:\mathrm{N}\right)=\mathrm{nj}_{\mathrm{2}} \left(\mathrm{mod}\:\mathrm{N}\right) \\ $$$$\:\:\:\alpha^{\mathrm{n}} =\beta^{\:\mathrm{m}} \:\Rightarrow\:\:\:\omega^{\:\mathrm{nj}_{\mathrm{1}} \left(\mathrm{mod}\:\mathrm{N}\right)} =\omega^{\:\mathrm{mj}_{\mathrm{2}} \left(\mathrm{mod}\:\mathrm{N}\right)} \Rightarrow\mathrm{nj}_{\mathrm{1}} \left(\mathrm{mod}\:\mathrm{N}\right)=\mathrm{mj}_{\mathrm{2}} \left(\mathrm{mod}\:\mathrm{N}\right) \\ $$$$\mathrm{Therefore}\: \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$$\mathrm{For}\:\:\:\:\:\:\:\:\:\:\:\mathrm{mj}_{\mathrm{1}} \left(\mathrm{mod}\:\frac{\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} }{\mathrm{k}}\:\right)=\mathrm{nj}_{\mathrm{2}} \left(\mathrm{mod}\:\frac{\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} }{\mathrm{k}}\:\right)\: \\ $$$$\:\mathrm{and}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{nj}_{\mathrm{1}} \left(\mathrm{mod}\:\frac{\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} }{\mathrm{k}}\:\right)=\mathrm{mj}_{\mathrm{2}} \left(\mathrm{mod}\:\frac{\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} }{\mathrm{k}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\:\alpha,\:\beta\right)=\left(\omega^{\mathrm{j}_{\mathrm{1}} } ,\:\omega^{\mathrm{j}_{\mathrm{2}} } \right)=\left(\omega^{\mathrm{j}_{\mathrm{2}} } \:,\:\omega^{\mathrm{j}_{\mathrm{1}} } \right) \\ $$$$\:\mathrm{where}\:\:\omega\:\:\mathrm{is}\:\:\left(\frac{\mathrm{m}^{\mathrm{2}} \:−\:\mathrm{n}^{\mathrm{2}} }{\mathrm{k}}\right)\:{th}\:\mathrm{root}\:\mathrm{of}\:\mathrm{unity}\left(\:\mathrm{with}\:\mathrm{m}\geqslant\mathrm{n}\right). \\ $$

Answered by Rasheed Soomro last updated on 17/Aug/15

Formula for number of pairs (𝛂,𝛃) fulfilling the conditions given in the question. ((m^2 −n^2 +k)/k) Since the number of (((m^2 −n^2 )/k))th roots is ((m^2 −n^2 )/k) and any one of them may be the value of α and to every α one value of β is attached. Hence number of (α,β) pairs is ((m^2 −n^2 )/k) . But (0,0) also satisfy the conditions (and this is not included in the roots) So ((m^2 −n^2 )/k)+1=((m^2 −n^2 +k)/k) .Remember that k=gcd(m,n) and m≥n.

$$\boldsymbol{\mathrm{Formula}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{number}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{pairs}}\:\left(\boldsymbol{\alpha},\boldsymbol{\beta}\right)\:\boldsymbol{\mathrm{fulfilling}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{conditions}} \\ $$$$\boldsymbol{\mathrm{given}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{question}}. \\ $$$$\frac{\boldsymbol{\mathrm{m}}^{\mathrm{2}} −\boldsymbol{\mathrm{n}}^{\mathrm{2}} +\boldsymbol{\mathrm{k}}}{\boldsymbol{\mathrm{k}}} \\ $$$$\boldsymbol{\mathrm{S}}\mathrm{ince}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\:\:\left(\frac{\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} }{\mathrm{k}}\right){th}\:\mathrm{roots}\:\mathrm{is}\:\frac{\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} }{\mathrm{k}} \\ $$$$\mathrm{and}\:\mathrm{any}\:\mathrm{one}\:\mathrm{of}\:\mathrm{them}\:\mathrm{may}\:\mathrm{be}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\alpha\:\mathrm{and}\:\mathrm{to}\:\mathrm{every}\:\alpha\:\mathrm{one} \\ $$$$\mathrm{value}\:\mathrm{of}\:\beta\:\:\mathrm{is}\:\mathrm{attached}. \\ $$$$\mathrm{Hence}\:\mathrm{number}\:\mathrm{of}\:\left(\alpha,\beta\right)\:\mathrm{pairs}\:\mathrm{is}\:\:\frac{\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} }{\mathrm{k}}\:.\:\mathrm{But}\:\left(\mathrm{0},\mathrm{0}\right)\:\mathrm{also}\:\mathrm{satisfy}\:\mathrm{the} \\ $$$$\mathrm{conditions}\:\left(\mathrm{and}\:\mathrm{this}\:\mathrm{is}\:\mathrm{not}\:\mathrm{included}\:\mathrm{in}\:\mathrm{the}\:\mathrm{roots}\right)\: \\ $$$$\mathrm{So}\:\:\:\frac{\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} }{\mathrm{k}}+\mathrm{1}=\frac{\boldsymbol{\mathrm{m}}^{\mathrm{2}} −\boldsymbol{\mathrm{n}}^{\mathrm{2}} +\boldsymbol{\mathrm{k}}}{\boldsymbol{\mathrm{k}}}\:.\mathrm{Remember}\:\mathrm{that}\:\mathrm{k}=\mathrm{gcd}\left(\mathrm{m},\mathrm{n}\right)\:\mathrm{and} \\ $$$$\mathrm{m}\geqslant\mathrm{n}. \\ $$

Commented by Rasheed Soomro last updated on 20/Aug/15

Who did like? Suppose liking by abc is more important for me then I would like to know about liker′s ID. So liking should be recorded with ID. PL demand for this. Of course Likers′ ID′s shouldn′t be there for full time but when we click.......We must know, ′′ Who did like? ′′

$$\boldsymbol{{Who}}\:\boldsymbol{{did}}\:\boldsymbol{{like}}?\:\boldsymbol{{Suppose}}\:\boldsymbol{{liking}}\:\boldsymbol{{by}}\:\boldsymbol{{abc}}\:\boldsymbol{{is}}\:\boldsymbol{{more}}\:\boldsymbol{{important}} \\ $$$$\boldsymbol{{for}}\:\boldsymbol{{me}}\:\boldsymbol{{then}}\:\boldsymbol{{I}}\:\boldsymbol{{would}}\:\boldsymbol{{like}}\:\boldsymbol{{to}}\:\boldsymbol{{know}}\:\boldsymbol{{about}}\:\boldsymbol{{liker}}'\boldsymbol{{s}}\:\:\boldsymbol{{ID}}. \\ $$$$\boldsymbol{{So}}\:\boldsymbol{{liking}}\:\boldsymbol{{should}}\:\boldsymbol{{be}}\:\boldsymbol{{recorded}}\:\boldsymbol{{with}}\:\boldsymbol{{ID}}. \\ $$$$\boldsymbol{\mathrm{PL}}\:\boldsymbol{\mathrm{demand}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{this}}. \\ $$$$ \\ $$$$\boldsymbol{{Of}}\:\boldsymbol{{course}}\:\boldsymbol{{Likers}}'\:\:\boldsymbol{{ID}}'\boldsymbol{{s}}\:\:\boldsymbol{{shouldn}}'\boldsymbol{{t}}\:\boldsymbol{{be}}\:\boldsymbol{{there}}\:\boldsymbol{{for}}\:\boldsymbol{{full}} \\ $$$$\boldsymbol{{time}}\:\boldsymbol{{but}}\:\boldsymbol{{when}}\:\boldsymbol{{we}}\:\boldsymbol{{click}}.......\boldsymbol{{We}}\:\boldsymbol{{must}}\:\boldsymbol{{know}},\:''\:\boldsymbol{{Who}} \\ $$$$\boldsymbol{{did}}\:\boldsymbol{{like}}?\:'' \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com